The discussion of \(\mathrm{RL}, \mathrm{RC},\) and \(\mathrm{RLC}\) circuits in this chapter has assumed a purely resistive resistor, one whose inductance and capacitance are exactly zero. While the capacitance of a resistor can generally be neglected, inductance is an intrinsic part of the resistor. Indeed, one of the most widely used resistors, the wire-wound resistor, is nothing but a solenoid made of highly resistive wire. Suppose a wire-wound resistor of unknown resistance is connected to a DC power supply. At a voltage of \(V=10.0 \mathrm{~V}\) across the resistor, the current through the resistor is 1.00 A. Next, the same resistor is connected to an AC power source providing \(V_{\mathrm{rms}}=10.0 \mathrm{~V}\) at a variable frequency. When the frequency is \(20.0 \mathrm{kHz},\) a current, \(I_{\mathrm{rms}}=0.800 \mathrm{~A},\) is measured through the resistor. a) Calculate the resistance of the resistor. b) Calculate the inductive reactance of the resistor. c) Calculate the inductance of the resistor. d) Calculate the frequency of the AC power source at which the inductive reactance of the resistor exceeds its resistance.

Step by step solution

01

Calculate the resistance of the resistor when connected to the DC power source

To calculate the resistance, we can use Ohm's law: \(V = IR\). We are given that the voltage across the resistor is \(V = 10.0\) V, and the current is \(I = 1.00\) A. Plugging in the values, we get: R = \(\frac{V}{I} = \frac{10.0}{1.00}\) = 10.0 \(\Omega\) So, the resistance of the resistor is 10.0 \(\Omega\).

02

Calculate the impedance of the resistor connected to the AC power source

The impedance, Z, of a resistor and inductor connected in series can be calculated using the formula: \(Z = \sqrt{R^2 + X_L^2}\) We have the resistance, R = 10.0 \(\Omega\), and the AC current, \(I_{\mathrm{rms}} = 0.800\) A. We also know that the AC voltage is \(V_{\mathrm{rms}} = 10.0\) V. Using Ohm's law for impedance, we get: \(Z = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} = \frac{10.0}{0.800} = 12.5\) \(\Omega\)

03

Calculate the inductive reactance of the resistor

Now that we have the impedance, Z, and resistance, R, we can calculate the inductive reactance, \(X_L\), using the equation: \(Z^2 = R^2 + X_L^2 \Rightarrow X_L = \sqrt{Z^2 - R^2} = \sqrt{12.5^2 - 10.0^2} = \sqrt{56.25} = 7.5\) \(\Omega\) So, the inductive reactance is 7.5 \(\Omega\).

04

Calculate the inductance of the resistor

We can find the inductance, L, using the formula for inductive reactance: \(X_L = 2\pi fL \Rightarrow L = \frac{X_L}{2\pi f}\) We are given the frequency, \(f = 20.0\) kHz = \(20.0 \times 10^3\) Hz. Plugging in the values, we get: \(L = \frac{7.5}{2\pi (20.0 \times 10^3)}\) = \(5.96 \times 10^{-6}\) H So, the inductance of the resistor is approximately \(5.96 \times 10^{-6}\) H.

05

Calculate the frequency at which the inductive reactance exceeds the resistance

We are asked to find the frequency at which the inductive reactance, \(X_L\), is equal to the resistance, R. Using the formula for inductive reactance, we can write: \(X_L = R \Rightarrow 2\pi f L = R \Rightarrow f = \frac{R}{2\pi L}\) Using the obtained values for R and L, we have: \(f = \frac{10.0}{2\pi (5.96 \times 10^{-6})} \approx 267.2\times 10^3\) Hz So, the frequency at which the inductive reactance of the resistor exceeds its resistance is approximately 267.2 kHz.

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